Kinetic vs Thermodynamic products: When is the tipping point?

Question

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1. What the question is really asking

A reaction can proceed along two competing pathways

[ \text{A}\;\xrightarrow{k_{1}}\;\text{B}\qquad\text{and}\qquad \text{A}\;\xrightleftharpoons[k_{-2}]{k_{2}}\;\text{C} ]

B is the kinetic product (it forms fast, but it is not the most stable).
C is the thermodynamic product (it is the most stable, but it may be formed more slowly because the forward step has a larger barrier).

The student wants a quantitative way to predict, for any temperature T, what fraction of the mixture is B and what fraction is C after a given reaction time t.
In other words we need a mathematical model that

uses measurable quantities – the activation free energies (or the Arrhenius parameters) for the three elementary steps (k_{1}, k_{2}, k_{-2}) – and
gives the product distribution (\displaystyle X_{B}(T,t),\;X_{C}(T,t)).

Below is a step‑by‑step derivation of such a model, followed by the final compact expressions that can be used directly.

2. Step‑by‑step derivation

2.1 Write the elementary rate laws

[ \begin{aligned} \frac{d[\mathrm A]}{dt} &= -k_{1}[\mathrm A]-k_{2}[\mathrm A]+k_{-2}[\mathrm C] \[4pt] \frac{d[\mathrm B]}{dt} &= k_{1}[\mathrm A] \[4pt] \frac{d[\mathrm C]}{dt} &= k_{2}[\mathrm A]-k_{-2}[\mathrm C] \end{aligned} \tag{1} ]

We assume the reaction mixture is well‑mixed, the temperature is constant, and the total concentration is small enough that the elementary rate constants are truly first order in the reacting species.

2.2 Express the rate constants as a function of temperature

For a first‑order elementary step the Eyring‑transition‑state expression (or the Arrhenius form) is most convenient:

[ k_i(T)=A_i\;e^{-\frac{E_{a,i}}{RT}} \qquad (i = 1,2,-2) \tag{2} ]

or, equivalently, with the activation free energy (\Delta G_i^{\ddagger}),

[ k_i(T)=\frac{k_{\mathrm B}T}{h}\;e^{-\frac{\Delta G_i^{\ddagger}}{RT}} \tag{2′} ]

All three (A_i) (or (\Delta G_i^{\ddagger})) are experimentally measurable (e.g., from a kinetic study at a single temperature).

The thermodynamic stability of C relative to A is described by the standard Gibbs energy change

[ \Delta G^\circ = -RT\ln K_{\mathrm{eq}} = -RT\ln\frac{k_{2}}{k_{-2}} \tag{3} ]

so that

[ k_{-2}=k_{2}\;e^{\frac{\Delta G^\circ}{RT}} . \tag{3′} ]

Equation (3′) is useful because it lets us replace the reverse rate constant by the forward one and the equilibrium constant (or (\Delta G^\circ)).

2.3 Solve the kinetic system

Equations (1) are linear with constant coefficients, therefore they can be solved analytically.
It is simplest to eliminate ([\mathrm A]) and ([\mathrm C]) by forming a 2×2 matrix for the A‑C sub‑system:

[ \frac{d}{dt} \begin{pmatrix} [\mathrm A]\[2pt] [\mathrm C] \end{pmatrix} = \underbrace{ \begin{pmatrix} -(k_{1}+k_{2}) & \;k_{-2}\[4pt] k_{2} & -k_{-2} \end{pmatrix} }_{\displaystyle \mathbf{M}} \begin{pmatrix} [\mathrm A]\[2pt] [\mathrm C] \end{pmatrix} \tag{4} ]

The eigenvalues (\lambda_{1,2}) of (\mathbf M) are

[ \lambda_{1,2}= -\frac{(k_{1}+k_{2}+k_{-2})}{2}\; \pm\; \frac{1}{2}\sqrt{(k_{1}+k_{2}+k_{-2})^{2}-4k_{1}k_{-2}} \tag{5} ]

Both eigenvalues are negative (the system is stable).
With the initial condition

[ [\mathrm A]_0 = C_0,\qquad [\mathrm B]_0=[\mathrm C]_0=0 \tag{6} ]

the solution of (4) can be written as a sum of two exponentials:

[ \begin{aligned} \mathrm A &= C_0\bigl( a_1 e^{\lambda_1 t}+a_2 e^{\lambda_2 t}\bigr)\[4pt] \mathrm C &= C_0\bigl( c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}\bigr) \end{aligned} \tag{7} ]

The coefficients (a_{1,2},c_{1,2}) are obtained by imposing (6) and are

[ \begin{aligned} a_1 &=\frac{\lambda_2 +k_{1}+k_{2}}{\lambda_2-\lambda_1}, \qquad & a_2 &=\frac{-\lambda_1 -k_{1}-k_{2}}{\lambda_2-\lambda_1}\[4pt] c_1 &=\frac{k_{2}}{\lambda_2-\lambda_1}, \qquad & c_2 &=-\frac{k_{2}}{\lambda_2-\lambda_1} \end{aligned} \tag{8} ]

(The algebra is straightforward; the interested reader can verify it by plugging (7)‑(8) into (4).)

Now integrate the B‑formation rate equation:

[ \frac{d[\mathrm B]}{dt}=k_{1}\mathrm A \quad\Longrightarrow\quad \mathrm B=k_{1}\int_{0}^{t}\mathrm A\,d\tau . \tag{9} ]

Carrying out the integral with (7) gives

[ \mathrm B=C_0\,k_{1} \Bigl( \frac{a_1}{\lambda_1}\bigl(1-e^{\lambda_1 t}\bigr) + \frac{a_2}{\lambda_2}\bigl(1-e^{\lambda_2 t}\bigr) \Bigr) . \tag{10} ]

2.4 Convert to mole fractions (product selectivity)

The total amount of material is conserved:

[ C_0 = \mathrm A+\mathrm B+\mathrm C. ]

Hence the fraction (or mole‑fraction) of each product at time (t) is

[ \boxed{ X_{B}(T,t)=\frac{\mathrm B}{C_0},\qquad X_{C}(T,t)=\frac{\mathrm C}{C_0},\qquad X_{A}(T,t)=1-X_{B}-X_{C} } \tag{11} ]

All the quantities in (11) are explicit functions of the temperature‑dependent rate constants through (2)–(3′).

2.5 Useful limiting forms

Situation	Approximation	Resulting product ratio
Very short reaction time ((t\ll 1/	\lambda_{1,2}	))	Expand the exponentials to first order: (e^{\lambda_i t}\approx 1+\lambda_i t)	(\displaystyle \frac{X_B}{X_C}\;\xrightarrow[t\to 0]{}\;\frac{k_{1}}{k_{2}}) (pure kinetic control)
Very long reaction time ((t\gg 1/	\lambda_{1,2}	))	All exponentials vanish ((e^{\lambda_i t}\to 0))	(\displaystyle \frac{X_C}{X_A}\;\xrightarrow[t\to\infty]{}\;K_{\mathrm{eq}}=\frac{k_{2}}{k_{-2}}=e^{-\Delta G^\circ/RT}) (pure thermodynamic control)
Intermediate time	Use full expressions (10)–(11)	Gives a smooth crossover that can be plotted versus T and t.

Thus the model predicts the crossover temperature (or time) at which the kinetic product ceases to dominate. One can locate the “tipping point

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